Problem: $\dfrac{ -p + 8q }{ 2 } = \dfrac{ -10p + 6r }{ 10 }$ Solve for $p$.
Explanation: Multiply both sides by the left denominator. $\dfrac{ -p + 8q }{ {2} } = \dfrac{ -10p + 6r }{ 10 }$ ${2} \cdot \dfrac{ -p + 8q }{ {2} } = {2} \cdot \dfrac{ -10p + 6r }{ 10 }$ $-p + 8q = {2} \cdot \dfrac { -10p + 6r }{ 10 }$ Multiply both sides by the right denominator. $-p + 8q = 2 \cdot \dfrac{ -10p + 6r }{ {10} }$ ${10} \cdot \left( -p + 8q \right) = {10} \cdot 2 \cdot \dfrac{ -10p + 6r }{ {10} }$ ${10} \cdot \left( -p + 8q \right) = 2 \cdot \left( -10p + 6r \right)$ Distribute both sides ${10} \cdot \left( -p + 8q \right) = {2} \cdot \left( -10p + 6r \right)$ $-{10}p + {80}q = -{20}p + {12}r$ Combine $p$ terms on the left. $-{10p} + 80q = -{20p} + 12r$ ${10p} + 80q = 12r$ Move the $q$ term to the right. $10p + {80q} = 12r$ $10p = 12r - {80q}$ Isolate $p$ by dividing both sides by its coefficient. ${10}p = 12r - 80q$ $p = \dfrac{ 12r - 80q }{ {10} }$ All of these terms are divisible by $2$ $p = \dfrac{ {6}r - {40}q }{ {5} }$